RElAtIoNs & FUnCtIoNs
LEcTuRe 1
z
Ith
A
1
b
2
C
3
d
4
A
B
basics
ke
muumuu
Two
sets
set
Input
:
,
output
set
is
AXB
t
Cartesian
product of
2 sets
→
set
output
t
input
set
here
each
element
every
For Ln
element
:
of
B
of
maps to
A
.
I
a
2
b
A
3
c
B
Y
=L 1.2.34 )
{ 13123133443
s
Sa b. e3
-
4,21 41,33
,
-
B
A
A- XB
is
{ (I
a
,
)
( Y at
,
Now
any
subset
for
,
of
En
,
d. b 1,11
c
)
( 2. a)
,
12
,
b)
,
12
,
c)
,(3.
at
,( 3 b)
,
,
4. b) 14,41
AXB
s
,
is
s cha )
,
Relation
a
(2.
a
)
,
( 3. all
from
is
A
a
to
B
.
RELATION
.
( 3. c)
.
In
•
•
In
12
Also
IM
has
unique
types of
Q, Set
A
R
}
I ( 2,41
=
In
R
P
In
Important
.
,
6,4 )
,
I
,
function
a
1)
called
is
Relations
me
But
.
Function
a
R
& P
dekhen.ge
input
if
relations
are
.
basic
knock
peke
use
such that
AXA
on
bar
questions
eete hai
-
,
/
12,31 14 , 1)
,
,
I 1
,
4,213
,
is
function ? ?
a
1 & 2
to
input
of
the
If
a
relation
no
input 14
ie
,
has
142
outputs
2
not
ie
defined
elements
of
.
1,2 , 3,4
A XB
on
A
of
only
has
then
if
,
it
,
output
L
the
P is
ie
,
has
relations
be
CANNOT
function
a
elements
more
function
a
.
.
Kyun ? ?
A
Dekho
case
number
input
bhi
me
mapped
How many
14
,
Each
:
set &
.
Par
Q,
(1,2 )
related
is
than
*
&
these relations
of
which
*
1,2 , 3 , 4
,
from input
element
is
n
,
relation
a
A XA
padhenge
,
y)
,
.
relations
( ( 2,1 ) (2
=
P
Are
{
is
ki
hum
me
form ( n
.
.
tha
output
multiple outputs
have
can
the
in
padha
me
relation
ke
And
R
from output set
is
'm
input
one
,
relation
a
y
•
relation
a
hai
agar Koi
function
&
of
set
relations
Agar
.
A
ho
relation
is
created
be
can
me
in
jisme
then
elements
defined from
elements
Saare
only
in
set A
to
hai
of
A
will enlist
pairs
of
manimum
m
are
.
?
elements
d
A-
No
of
.
No
.
of
Now
So
,
,
relations
of
.
It
no
subsets
.
Proof [No of
.
of
of
.
AXA
elements in
no
no
=
=
mxm
2K ( where
=
relations
subsets
subsets
in
a
m2
=
:
no
=
possible
are
set ]
AXA
of
k
AXA
P
{
=
a , .az
.
Moth C , th
Cz th Cz
"
.
.
.
→
20h
"
Cn
-
in
that set )
.
{ a } 1923
→
elements
of
.
E
'
I
'
.
.
.
.
-
.
.
an }
La .az } Ear 933
San }
,
n
subsets
C,
)
{
a
,
,
Az
,
0373
h
Cz
.
Cz
.
-
.
.
.
.
KEELE
If
:
Relations
of
Types
R
is
then
relation
a
it is
For En :
known
A
if
R
&
R
*
Every
If
A
has
If
:
elements
for
R
is
present
it
.
12,21
,
R
R
13,3 ) )
,
4,21
,
12,2 ) )
,
s
n
ER
)
is
t
other
Cn
&
to
y
,
F
EA
n
13,3 ) }
,
→
R
.
is
reflexive
.
.
pairs ( not nessecary ]
has minimum
AXA
on
of
.
also be
Cy , n ) should
then
symmetric
be
reflexive
reflexive
R
E R
l
)
l
,
not
Relation
any
R is
then
R
REFLEXIVE
it
,
.
621,13
,
itself
with
then
,
for
)
I
,
relation
if
-
11
,
relation
(n
31
,
to be
a
in
For En
Ill
=
mapped
is
m
elements
m
fyymEEk
element
{ It
=
R
1,2
{ 4,11
=
&
AXA
reflexive
as
I
=
the set
on
.
{ 1,2 3,41
A
is
=
{ ( 1,21 ( 2,1 I
,
,
.
(l
,
I
)
)
Symmetric
is
.
-
Unlike
with
R
*
:
symmetric
is
not nesseciary
But
In y )
it
,
element
an
is
If
Transitive
reflenive
.
if
any
,
that
d ly , a ) is
present
is
of
each element
A
cis
present
mapped
then
.
n
is
related
If
in
a
n
should be
to
then
y
R
relation
n
,
related to
related
is
for
z
to
also related
is
y
to
y
to be
it
a
.
y is related to
,
transitive
z
then
.
or
*
if
(n
For 9N
:
,
A
Cy
&
y ) ER
{
=
R
=
R
=
,
2) ER
I, 2
,
{ ( 1,2 )
{ (4,21
then
3,
43
,
(2,31
,
( 2,3 )
Cn, 2) ER
( 1,3)
,
,
4,21
,
,
for transitivity
)
e
11,3 ) }
→
l l,
4
)
is
.
Transitive
Not
Transitive
*
A relation which
is
reflexive
o
,
symmetric
& transitive
is
called
an
Equivalence Relation
.
Some questions
Tyke
-
:
A
Let A
=
L
:
& R
given
are
mm
0=1
R
{
=
10
10,0 )
33
1,2 ,
,
( 0,11
,
Is R reflenive ?
ALI
( 0,01
:
(0,1 )
11,1 )
,
for 10,01
for
(O
for
(
for
Cl ,
for
11,0 )
Pls
A- 2
-
Am-2 :
B
.
has
is
&
( 0,3 )
-
,
l)
,
31
12,2 )
( 1,11
,
only
(0,3)
(
present
I
is
,
is
is
for practice
relation
Refleniue
is
R
ie
,
is
.
Symmetric
.
.
ie
R is
not
A
follows
Transitive
.
.
R in the set
as
:
.
added
to
R
to
make it
already
ie
it
is
Reflexive
be
to
pairs
in R
present
not
define
11,3 ) }
R
ie
present
1
,
.
EXEMPLAR ]
present
Enercise
a
.
( NCERT
in R
:
present
is
4,3 )
13,0 )
,
follows
13,313
as
present
is
O)
A
on
present
are
1
I
(I , I )
→
,
the
smallest equivalence
EXEMPLAR]
(2,2 )
(2,21
13,1)
( 3,1 )
,
( 3,3 )
(3,1 )
but
,
&
→
13,3 )
,
ordered
for 11,3 )
(O ,
→
-
1,2 ,
{ ( 1st )
→
→
(O
R has ( 1,3 )
So
)
12,2 )
.
Transitive ?
(3,0 )
→
(0,0 )
( NCERT
.
Mr
R
)
d
-
I
=
the
relation
l
l
,
(0,3
) &
O
E 11,11
Write
(O
4
N C E R T
solve
=
10,3 )
,
11,11
,
are
,
R
relation
a
I 1,01
,
(2,21 13,3 )
,
1,0 ) &
For set A
R
O
,
10,3 )
Symmetric ?
11,01
→
define
&
,
:
,
s
missing
ill
3) 13,111
( 3,3 )
is
,
( 2,1 )
should
be
so
,
,
is
present
added in
add
13
,
I
)
.
.
Now it is
it
is
Transitive
R
to
make it
also
symmetric
as
.
Equivalence
.
also
.
Type
2
-
R
:
I
=
hmmm
3
T be
Let
R
AIL
:
=/
:
,
set
:
T , is
,
Some
relation
b/w
all
triangles
in
of
the
CT Tz )
,
In y )
congruent
R
is
Also
is
symmetric bcz
R
to T.
congruent
R in
transitive
then
9,4 Show
definitely
R in
R
is
Am -4
:
let
01,5
{
=
is
reflexive
since
R
vis
symmetric
because
R
vis
transitive
we
can
ie
a
ie
R
R be
Then R
write
is
if
the relation
on
,
a)
T, ) ER
Tz
to
Tz
.
also
is
.
& Tz
Tz
to
congruent
is
of integers given by
divides a b )
-
-
if
a
2
by
-
2 divides
b
b
d
m
,
-
C
=
-
b
a
-
then
divisible
is
it
by
2n
a
will
&
2
-
0
2 divides
or
c
=
-
is
c
(a
-
b )
+
(b
-
C
)
=
(c)
N
natural numbers
of
defined by
n
Rm
if
n
Equivalence
(D) Reflexive
Ans S
-
:
R
( (n
=
nm
R is
is
R
is
ie
,
m
)
:
reflexive
R
n
transitive but not
,
not
since
Symettnc
Transitive
divides
(D )
divides
n
p
is
also
as
.
symmetric
as
if
EXEMPLAR]
( NCERT
.
)
always
divide
2 divides 4
but
will
n
m
.
correct
m
u
=
a
.
2
21Mt n )
.
Symmetric
and
-
by
divisible
vis
Transitive
b
divide
definitely
b
.
.
relation
the set
a
a
(A) Reflexive and Symmetric
(B)
by
(NCERT)
.
divides
2
equivalence
an
,
then
itself
.
[NCERT]
.
2
=
divisible
is
c
-
b
-
-6
Ta
given
,
relation
to
congruent
is
to
equivalence
an
T
relation in
a
Z
2
:
,
because
a
to Tz
la b )
for Ca
R
mm
congruent
is
the set
relation
equivalence
an
IT
T,
as
then
be
R is
congruent
T , is
congruent
is
the relation
that
if
ER
.
given }
& R
Show that
.
Tz ) ER
, ,
as
T,
if
if (T
ie
.
,
is
plane
a
CT Ti )
reflexive by
clearly
}
to Tz
dy
n
divides
in
,
d
m
p
=
n
4
m
Mm
itself
.
doesn't
divides
→
divide
p
then
p =Cµ Hn
I
2
obviously
divides
me
.
RElAtIoNs & FUnCtIoNs
LEcTuRe 2
Functions
Already
last lecture
note hair
untie
CE If R
is
range
E
elements
set
create
&
lecture
One
C
3
d
4
A
B
ki
tha
bolte hair
R
,
( ( 1,21
=
One
-
input
from
range
ie
1)
2
relations
vo
inputs
jisme
ke
UNIQUE
outputs
.
(2,3 )
,
,
(3,31 (4,211
,
.
Find its domain &
.
Here
Ab is
b
dekha
me
function
a
1
:
Functions
mum
A
is
:
Range
is
:
Functions
(#)
a
2
b
make the domain
{ 1,2
I
2,
,
3 Y
,
& elements
from output
)
33
types of functions padhenge
hum
1
together
.
Domain
me
set
.
.
:
theft
input
a
different
.
3
C
No two
Y
d
one
A
B
-
has
one
inputs have same image
functions are called injective
.
Functions
.
Q,
"
check whether
f-
Ill
is
2
-
inputs
mmmm
a
a
,
)
,
L2
one
,
same
s
Definition
Mathematical
In
is
f
b)
-
13
,
,
a
not ?
or
one
)
output
.
Hi d) 3
Is
→
3
a
,
function
this
a
→
is not one
-
ore
.
:
i m
function f
:
for
As B
t
any
n ,
,
E A
na
f-In )
,
,
t flu )
is
not
&
equate f- In )
.
/
domain
co domain
-
If any
A
two
to
way
inputs give
prove
in
the
output
same
questions
take
:
it
if
ie
if
Graphical interpretation
Hn )
=
n
,
y
C-
Many
-
one
functions
:
A
1- a
3
/
Y
A
that
fly )
n
=
n
then
y
-
-
then
y
f
one
is
-
one
fin)
one
-
.
fly )
&
is
one
one
-
,
one
.
.
:
crummier
2
Domain
concludes
/
2)
function
then
b
c
d
B
function
~
which is
not
Here
both
ie
,
two
one
-
one
2 &
inputs
3
have
known
is
are
the
as
a
mapped
to
same
output
Many
b
.
.
-
one
function
.
f- In )
Oi, Prove that
¥
For
one
one
-
is
a
y E Domain
n,
:
7
at
=
one
f- In )
e
n
n
ie
flat
Q, Prove that
¥
For
one
:
one
-
n'
=
n
,
9
-
f
is
not
is
y
-
-
one
flat
y ED
one
-
one
-
-
fly )
=
of-9
y -17
=
one
.
fly )
=
-17
function
one
-
y
=
'
9
-
n2=y2
n
Also
fll )
,
R 9
=
=
-
-
,
fl
is
not
8
ffQ,
A
function
Is
As
f
i
1
2
→
{
I
33
2,
,
-
Q, How many
one
&
B
a)
if
man
b)
if
m
c)
if
m > n
if
&
f
1-
L
**
As
f- IT 9
=
one
many
-
=
-
-
8
-
one
one
A
where
Gl , 2,33
vis
A
no
has
n
-
Range
I
→
1 , 2,3
.
(2,3 ) (3 I ) )
,
,
}
has
-
one
is
functions
,
one
-
injective
is
elements
-
one
can
be
defined from
that
considering
every
A
to
B
if
element
of
A
is
elements
m
,
other element will be
has
I
n
-
l
or
l
"
so
pair
each element will
then
mapped
on
.
So
no
with it
.
.
of ways
The
:
with
n
-
Cn
Pm
-
m
A has
taken
=
mon
n
then
then
"
no
Pn
=
one
-
n
one
!
function
can
exist
.
-
exactly
element
first
Ln ml !
if
{ ( 1,2 )
is
m
elements
.
n
the neut
if
f
:
13
2
*
)
AXA
defined from
is
t
-
injective ?
Domain
3
is
ty
=
t
)
.
.
.
has
n
-
(m
&
element
one
choices
n
-
l
l
,
Also
ie
,
Range of flu)
f
A, Show that
E clearly
& also
to be
Onto
is
one
A-
that
the
y
=
Un +5
=
codomain
then
onto
must
be
:
l l 2,3 }
,
e
output
each
mapped
to
one
f
l,
2,33
must be
&
only
is
always
one
mapped from
one
output
.
So
.
coexecutor Pack
Q, Show
function f
the onto
-
bcz
.
.
if f is
each input
one
R
is
:
one
-
one
function f
l :#
onto
:
El 2,33
,
our
→
L
l,
2,33
is
onto
41*981
.
-
one
atleast
the
.
one
function
input
has